I was reading Andrew Broker's paper on his breakthrough with the sum of three cubes problem. In it, I was confused by one result. Earlier in the paper, it is stated that:

$$ x^3 + y^3 + z^3 = k \mid k,x,y,z \in \mathbb{Z} \ni k \equiv 3 \textrm{ (mod 9)} \implies x \equiv y \equiv z \equiv 1 \textrm{ (mod 3)}$$

Which totally makes sense to me. However, later, at the very end, it states in the explicit case of ($k=3$) with $|x| \geq |y| \geq |z|$ without loss of generality, it follows that:

$$ z \equiv \frac{4(x+y)^2}{3} + 3((x+y)^2-1) \textrm{ (mod 162)}$$

However, I do not see how $z \equiv 1 \textrm{ (mod 3)}$ is ever possible then. For $\frac{4(x+y)^2}{3}$ to be integral, 3 must divide $(x+y)$, which implies that $\frac{4(x+y)^2}{3} = 3k$ for some integer $k$. Meanwhile, $3((x+y)^2-1)$ is trivially a multiple of 3. Since this is under a modulus which is a multiple of three, wouldn't this imply that $z$ must also be a multiple of three, contradicting the first result? They did a computer search for solutions, so I assume I'm missing something, but I don't see what.

There are two small, previously found solutions, (1,1,1 & -5,4,4), but for neither part does the second claim hold true. I would presume it is only for larger solutions, or perhaps solutions where $x > y > z$, but at this point I don't know exactly what's going on.

So, in summary, why did they search for primes if $z$ cannot be 1 mod 3? Did I misread something or what?

  • $\begingroup$ As for the solutions $(1,1,1)$ and $(-5,4,4)$, they assume that $y\ne z$ in the beginning of the second part. $\endgroup$ – Milten Mar 28 at 11:20
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    $\begingroup$ You misunderstood the congruence as it is $4(\frac{x+y}{3})(x+y)$ and the first term is a reciprocity symbol, not a division, so $x+y$ actually must be coprime with $3$ as assumed $\endgroup$ – Conrad Mar 28 at 12:09
  • $\begingroup$ apologies, what is a reciprocity symbol? $\endgroup$ – Zachary Hunter Mar 28 at 13:35
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    $\begingroup$ For an odd prime $p$ and any integer $a$, $(a/p)=0$ if $p$ divides $a$. If not: $(a/p)=1$ if there are solutions to $x^2\equiv a \pmod p$, else it equals $-1$. i.e. whether you can take squareroot modulo $p$. Also see Legendre symbol $\endgroup$ – Yong Hao Ng Mar 28 at 13:50

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